Position for maximum electric field between two wires

[PhysicsKush]

Homework Statement

Given a two-conductor setup of two infinitely long parallel wires. Let the conducting wires have radius $R$ and be separated by a distance $D$, where $D$ is much larger than $R$. Use the approximation that the field from the second wire does not impact the charge distribution on the first. Similarly, you may assume that the field near a wire is dominated by that wire and you can safely neglect the field from the other wire, and that the wire diameter is small compared to the separation. Where then is the electric field going to be maximum ? What is the potential difference between the two wires, expressed in terms of $\lambda$ ?

Relevant Equations

For a wire
$$V = \frac{\lambda}{2 \pi \epsilon_{0}} \ln \Bigg\lvert \frac{r}{a} \Bigg\rvert,$$
where $r$ is the distance to the point and $a$ is the distance from the reference point.
$$ E = \frac{\lambda}{2 \pi \epsilon_{0} r},$$
where $r$ is the radius of the wire.

For the first part, since
$$ E(r) \propto \frac{1}{r} \hat{r}$$
by the principle of superposition the maximal electric field should be halfway in between the two wires.

Then I'm not sure how to go about the second part of the question. I understand that the total potential due to the two wires on an $2$ dimensional plane , at an arbitrary point is
$$ V = \frac{\lambda}{2 \pi \epsilon_{0}} \ln\Bigg\lvert\frac{d_{-}}{d_{+}}\Bigg\rvert,$$
where $d_{-} ,d_{+}$ are respectively the distances between each wire to the evaluated point. I also know that the potential difference between two points is
$$ V(b) - V(a) = \int_{a}^{b} \vec{E} \cdot d\vec{l},$$
but I'm unable to translate these informations into a correct answer.

Any insight would be appreciated

 

[Abhishek11235]

Let $d_- \rightarrow d$ (The distance between 2 wires) and $d_+ \rightarrow a$ (The radius of wire) in the formula of potential you wrote.

 

[vela]

$$ E = \frac{\lambda}{2 \pi \epsilon_{0} r},$$ where $r$ is the radius of the wire.

That's not what $r$ represents.

For the first part, since
$$ E(r) \propto \frac{1}{r} \hat{r}$$
by the principle of superposition the maximal electric field should be halfway in between the two wires.

How did you come to this conclusion? I'd expect the field to vanish halfway between the two wires.

 

[PhysicsKush]

That's not what $r$ represents.How did you come to this conclusion? I'd expect the field to vanish halfway between the two wires.

You're right, well I think $r$ is the Gaussian radius which can be extended over the real radius, so essentially it is the distance from the center of the wire.

Also, well since both wire produce an electric field that decays with $\propto \hat{r}/r$, oh wait I realize my answer makes no sense ! There should be two points where the maximal electric field should occur : right near the center of each wire ?

 

[vela]

Also, well since both wire produce an electric field that decays with $\propto \hat{r}/r$, oh wait I realize my answer makes no sense ! There should be two points where the maximal electric field should occur : right near the center of each wire ?

Remember that the wires are conductors.

 

[PhysicsKush]

Remember that the wires are conductors.

Right , the electric field can not be near the center because the electric field inside the conductor is $0$ ! I conclude the maximal electric field occurs right over the surface of the wire ?

 

[haruspex]

Is λ the charge density for both? Why would there be a potential difference? Or is one of them -λ?

 

[PhysicsKush]

Is λ the charge density for both? Why would there be a potential difference? Or is one of them -λ?

It is not stated, but I can only assume that is the logical case.

 

[PhysicsKush]

So I digged further and found out that by superposition the total electric field between the two wires should be
$$ \frac{\lambda}{2\pi \epsilon_{0}R} + \frac{\lambda}{2\pi \epsilon_{0} (d-R)} = \frac{\lambda}{2 \pi \epsilon_{0}}\left( \frac{d}{R(d-R)}\right),$$
since $R \in (0,d)$, it follows the maximal electric field is right over the surface of each wire in the direction towards the opposite wire ,i.e., at $R = 0^{+}$ and $R=d^{-}$

What do you guys think? I'm not sure this makes sense because in a subsequent question I have to manipulate mathematically the maximal value of $E$, how do I maximize $\frac{d}{R(d-R)}$?

 

[guv]

$R(d-R)$ is a parabola opening downwards, I do not see how you would be able to maximize the fraction.

 

[vela]

So I digged further and found out that by superposition the total electric field between the two wires should be
$$ \frac{\lambda}{2\pi \epsilon_{0}R} + \frac{\lambda}{2\pi \epsilon_{0} (d-R)} = \frac{\lambda}{2 \pi \epsilon_{0}}\left( \frac{d}{R(d-R)}\right),$$
since $R \in (0,d)$, it follows the maximal electric field is right over the surface of each wire in the direction towards the opposite wire ,i.e., at $R = 0^{+}$ and $R=d^{-}$

What is $d$? $R$ is the radius of the wire. It doesn't make sense to say $R = 0^{+}$ and $R=d^{-}$. How come the distance $D$ between the wires doesn't appear in this answer?

The first thing you should do, though, is figure out what the charge on each wire is. Do they both have charge density $\lambda$, or does one have $\lambda$ and the other $-\lambda$? It seems pointless to continue until you can give us a clear statement of the problem.